Question

L L2 L L 7. [1 pt) The truss is pin-joined to the wall. Find the force in each member. Assume L, = 2.71 m, L 1.83 m, P = 1024 N. If the member is in compression, enter a negative force. Enter FAB Answer: Sube lewe 8. [1pt) Enter FBC Answer: Submit Al Answers 9. [1 pt) Enter FBD Answer: 10. [1pt] Enter FCD Submit Alwers Answer: Submit Al Answers 11. [1pt] Enter FDE Submit Al Answers Answer: 12. [1 pt) Enter FAD Submit Alwe Answer: 13. [1pt] What is the greatest force P that can be applied if the forces in any of the members are not to exceed 2290 N in tension or 1810 N in compression? Answer: Submit Al www

Answer 1

solution 0 x Given len D А Cy P= 1024N L = = 2.71m h Le = 1.83m B EF = 0 Co > Excel Pg=o7 Ey-cy = 1024N MBC = o → — (1024x2712) + Erex (1-83) C) = 0 Ex = 1516.4153 N x= 1516.4153N use method of joints (consider all members as Tension) o o = tant (1.83 34:03° E Ex FED Efn cos Ex= En Coso ED Ey ED - 1829.7738N Ify 20→ Ey FED sino Fy 1023.9907 N from @ Cy -9.3x10 3N [Cyr on

o = tan! Eco (1.83 2:71 Oq = tani tani/1-83 = 34.03 c Cxor 2.71 < PCB co EFG =0 = - Ep Coso, - FEB coso, - Сх {Fy=0) Eosino, = FEB sin Oz FED = FB - (Eco Corles) e) =C FED - 914.8869N FBE 914.886 9N 7 EF=09 34.03° DE 34.03 of Cos(34•03) + Fe Cos(34:03)+E, Cos(34-03) Fco A - FDA 'DA 914.8869N 34.03 FED EFq zo - Fept Sin 24-03) – Fosine4.03) -Forvin B4.03)20 = 1023.9806N FBD
6
3 DA 7 34003 50 FoA Cos(34.03 )+ Fea Cos(34.03) = 0 34:03 FBA -914.8869 N 1024 E ВА Hence 7 FAB = -89148869N 8) FBC -914.8869N १) 1023.9806N BD Eco = - 914 8869N 11) FDE 1829.7738N 12) FAD =914.8869 N 13 'AD A {fy zo Far Sin (34.03) + Pre Sin34.03 ) = mae АВ 'Pmax Given Pmax E2294.4703N = 2990 N AD FOR =1810N